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ECE Ph.D. Qualifying Exam in "Communication, Networks, Signal, and Image Processing" (CS)

Question 1, August 2011, Part 1

Part 1,2]

 $ \color{blue}\text{1. } \left( \text{25 pts} \right) \text{ Let X, Y, and Z be three jointly distributed random variables with joint pdf} f_{XYZ}\left ( x,y,z \right )= \frac{3z^{2}}{7\sqrt[]{2\pi}}e^{-zy} exp \left [ -\frac{1}{2}\left ( \frac{x-y}{z}\right )^{2} \right ] \cdot 1_{\left[0,\infty \right )}\left(y \right )\cdot1_{\left[1,2 \right]} \left ( z \right) $

$ \color{blue}\left( \text{a} \right) \text{ Find the joint probability density function } f_{YZ}(y,z). $

$ \color{blue}\text{Solution 1:} $

$ f_{YZ}\left (y,z \right )=\int_{-\infty}^{+\infty}f_{XYZ}\left(x,y,z \right )dx $

$ =\frac{3z^{2}}{7\sqrt[]{2\pi}}e^{-zy}\int_{-\infty}^{+\infty}exp\left[-\frac{1}{2}\left(\frac{x-y}{z} \right )^{2} \right ]dx\cdot 1_{[0,\infty)} \left(y \right )\cdot1_{\left [1,2 \right ]}\left(z \right ) $

$ \text{But}\int_{-\infty}^{+\infty}exp\left[-\frac{1}{2}\left(\frac{x-y}{z} \right )^{2} \right ]dx \text{, looks like the Gaussian pdf, so} $

$ =\frac{3z^{2}}{7\sqrt[]{2\pi}}e^{-zy} \underset{{\color{Orange}\sqrt[]{2\pi}z}} {\underbrace{ {\color{Orange} \frac{7\sqrt[]{2\pi}z}{7\sqrt[]{2\pi}z} } \int_{-\infty}^{+\infty}exp\left[-\frac{1}{2}\left(\frac{x-y}{z} \right )^{2} \right ]dx}}\cdot 1_{[0,\infty)} \left(y \right )\cdot1_{\left [1,2 \right ]}\left(z \right ) $

$ =\frac{3z^{2}}{7}e^{-zy}\cdot 1_{[0,\infty)} \left(y \right )\cdot1_{\left [1,2 \right ]}\left(z \right ) $

$ \color{blue}\text{Solution 2:} $

$ f_{YZ}(y,z) = \int_{-\infty}^{\infty}{f_{XYZ}(x,y,z)dx} $

$ = \int_{-\infty}^{\infty}{\frac{3z^2}{7\sqrt[]{2\pi}} e^{-zy} \cdot e^{-\frac{1}{2} \frac{(x-y)^2}{z^2}} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z) dx} $

$ = \int_{-\infty}^{\infty}{\frac{1}{7\sqrt[]{2\pi}z} e^{-\frac{(x-y)^2}{2z^2}} \cdot \frac{3}{7}z^3 e^{-zy} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z) dx} \color{green}\text{ Here the student wants to form a Gaussian pdf.} $

$ \color{green}\text{Based on the Axioms of Probability, this integral over R will be 1.} $

$ \color{green}\text{So he wants to replace this integral with 1:} $

$ = \frac{3}{7}z^2 e^{-zy} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z) $

$ \color{blue}\left( \text{b} \right) \text{Find } f_{x}\left( x|y,z\right ) $

$ \color{blue}\text{Solution 1:} $

$ \color{green}\text{According to the Bayes rule:} $

$ f_X(x|y,z) = \frac{f_{XYZ}\left( x,y,z\right )}{f_{YZ}\left(y,z \right )} $

$ = \frac{e^{-\frac{1}{2}\left(\frac{x-y}{z} \right )^{2}}}{\sqrt[]{2\pi}z} $  

$ \color{blue}\text{Solution 2:} $

$ f_X(x|y,z) = \frac{f_{XYZ}(x,y,z)}{f_{YZ}(y,z)} = \frac{\frac{3z^2}{7\sqrt[]{2\pi}} e^{-zy} \cdot e^{- \frac{(x-y)^2}{2z^2}} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z)} {\frac{3}{7} z^3 e^{-zy} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z)} $

$ \frac{1}{\sqrt[]{2\pi}z} e^{- \frac{(x-y)^2}{2z^2}} $

$ \color{blue}\left( \text{c} \right) \text{Find } f_{Z}\left( z\right ) $

$ \color{blue}\text{Solution 1:} $

$ f_Z(z) = \int_{0}^{+\infty}{f_{YZ}\left(y,z \right )dy} $

$ =\frac{3z^{2}}{7}\cdot1_{\left[1,2 \right ]}(z) $  

$ \color{blue}\text{Solution 2:} $

$ f_Z(z) = \int_{-\infty}^{\infty}{f_{YZ}(y,z)dy} $

$ = \int_{0}^{\infty}{\frac{3z^3}{7} e^{-zy} \cdot 1_{[1,2]}(z) dy} $

$ = \frac{3z^2}{7} \cdot \int_{0}^{\infty} z e^{-zy} dy \cdot 1_{[1,2]}(z) $

$ = -\frac{3z^2}{7} \cdot e^{-zy} |_{0}^{\infty} \cdot 1_{[1,2]}(z) $

$ = \frac{3}{7} z^2 \cdot 1_{[1,2]}(z) $

$ \color{blue}\left( \text{d} \right) \text{Find } f_{Y}\left(y|z \right ) $

$ \color{blue}\text{Solution 1:} $

$ f_Y(y|z) = \frac{f_{YZ}\left(y,z \right )}{f_{Z}(z)} $

$ =e^{-zy}z\cdot1_{\left[0,\infty \right )}(y) $  

$ \color{blue}\text{Solution 2:} $

$ f_Y(y|z) = \frac{f_{YZ}(y,z)}{f_Z(z)} = \frac{\frac{3}{7} z^3 e^{-zy} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z)}{\frac{3}{7} z^2 \cdot 1_{[1,2]}(z)} $

$ = ze^{-zy} \cdot 1_{[0,\infty)}(y) $

$ \color{blue}\left( \text{e} \right) \text{Find } f_{XY}\left(x,y|z \right ) $

$ \color{blue}\text{Solution 1:} $

$ f_{XY}(x,y|z) = \frac{f_{XYZ}\left(x,y,z \right )}{f_{Z}(z)} $

$ =\frac{e^{-zy}}{\sqrt[]{2\pi}}e^{-\frac{1}{2}\left(\frac{x-y}{z} \right )^{2}}\cdot1_{\left[0,\infty \right )}(y) $  

$ \color{blue}\text{Solution 2:} $

$ f_{XY}(x,y|z) = \frac{f_{XYZ}(x,y,z)}{f_Z(z)} $

$ = \frac{\frac{3z^2}{7\sqrt[]{2\pi}} e^{-zy} \cdot e^{-\frac{1}{2} (\frac{x-y}{z})^2} \cdot 1_{[0,\infty)}(y) \cdot 1_{[1,2]}(z)}{\frac{3}{7} z^2 \cdot 1_{[1,2]}(z)} $

$ = \frac{1}{\sqrt[]{2\pi}} e^{-zy} \cdot e^{-\frac{1}{2} (\frac{x-y}{z})^2} \cdot 1_{[0,\infty)}(y) $

"Communication, Networks, Signal, and Image Processing" (CS)- Question 1, August 2011

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• Part 1: solutions and discussions
• Part 2: solutions and discussions

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